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goldfiish [28.3K]

3 years ago

14

Solve the system of equations below by graphing both equations with a pencil and paper. What is the solution? y = x+1 1 X+ 3 2.

O A. (-4,-3) OB. (-2,1) O C. (2,3) O D. (4,5)

1 answer:

Troyanec [42]3 years ago

8 0

Answer:

D. (4, 5)

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS
Equality Properties

<u>Algebra I</u>

Solving systems of equations using substitution/elimination
Solving systems of equations by graphing

Step-by-step explanation:

<u>Step 1: Define systems</u>

y = x + 1

y = 1/2x + 3

<u>Step 2: Solve for</u><u><em> x</em></u>

Substitute in <em>y</em>: x + 1 = 1/2x + 3
Subtract 1/2x on both sides: 1/2x + 1 = 3
Subtract 1 on both sides: 1/2x = 2
Divide both sides by 1/2: x = 4

<u>Step 3: Solve for </u><em><u>y</u></em>

Define original equation: y = x + 1
Substitute in <em>x</em>: y = 4 + 1
Add: y = 5

<u>Step 4: Graph</u>

<em>We can confirm our answer.</em>

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The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s

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The positions when the particle reverses direction are:

$s(t_1)=55ft\\\\s(t_2)=28ft$

The acceleraton of the paticle when reverses direction is:

$a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}$

Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

$s(t)=2t^{3}-21t^{2}+60t+3$

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- Derivating to get the velocity function, we have:

$v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60$

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

$v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s$

We know that the particle reversed its direction two times.

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$a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42$

Now, substituting the times to calculate the accelerations, we have:

$a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}$

Now, substitutitng the times to calculate the positions, we have:

$s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft$

Have a nice day!

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